Water is flowing with a velocity of 2 m/s in | vedclass

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(C) Using the equation of continuity: $A_1 V_1 = A_2 V_2$. Given $A_1 = 2 	imes 10^{-2} \ m^2$,$V_1 = 2 \ m/s$,and $A_2 = 0.01 \ m^2 = 10^{-2} \ m^2$

Vedclass · Accepted Answer

$3.4 	imes 10^4$

Vedclass · Answer

(C) Using the equation of continuity: $A_1 V_1 = A_2 V_2$. Given $A_1 = 2 	imes 10^{-2} \ m^2$,$V_1 = 2 \ m/s$,and $A_2 = 0.01 \ m^2 = 10^{-2} \ m^2$. $(2 	imes 10^{-2})(2) = (10^{-2}) V_2$. $V_2 = 4 \ m/s$. Using Bernoulli's equation for a horizontal pipe $(h_1 = h_2)$: $P_1 + \frac{1}{2} ho V_1^2 = P_2 + \frac{1}{2} ho V_2^2$. Taking density of water $ho = 10^3 \ kg/m^3$. $4 	imes 10^4 + \frac{1}{2}(10^3)(2)^2 = P_2 + \frac{1}{2}(10^3)(4)^2$. $40000 + 2000 = P_2 + 8000$. $P_2 = 42000 - 8000 = 34000 \ Pa = 3.4 	imes 10^4 \ Pa$.

Water is flowing with a velocity of $2 \ m/s$ in a horizontal pipe where the cross-sectional area is $2 \times 10^{-2} \ m^2$ at a pressure of $4 \times 10^4 \ Pa$. The pressure at a cross-section of area $0.01 \ m^2$ in Pascal will be:

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